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  • Hdu 2602 Bone Collector

    Bone Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 61876    Accepted Submission(s): 25787

    Problem Description

    Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
    The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

     

    Input

    The first line contain a integer T , the number of cases.
    Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

    Output

    One integer per line representing the maximum of the total value (this number will be less than 231).

    Sample Input

    1

    5 10

    1 2 3 4 5

    5 4 3 2 1

    Sample Output

    14

    #include<stdio.h>
    #include<string.h>
    #define max(a,b) a>b?a:b
    int N,V;
    int dp[1050];
    int vol[1050],val[1050];
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d %d",&N,&V);
            for(int i=1;i<=N;i++)
                scanf("%d",&val[i]);
            for(int j=1;j<=N;j++)
                scanf("%d",&vol[j]);
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=N;i++)
                for(int v=V;v>=vol[i];v--)
                    dp[v]=max(dp[v],dp[v-vol[i]]+val[i]);
            printf("%d
    ",dp[V]);
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangliu/p/7063211.html
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